Question

Consider a linear transformation $\mathit{L}\left(\stackrel{\rightharpoonup }{x}\right)\mathbf{=}\mathit{A}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$fromlocalid="1662031612255">ℝnto localid="1662031573789">ℝm, with $\mathit{k}\mathit{e}\mathit{r}\left(L\right)\mathbf{=}\left\{\stackrel{\rightharpoonup }{0}\right\}$. The pseudoinverse L⁺ of L is the transformation fromlocalid="1662031665028">ℝmto localid="1662031688923">ℝngiven by ${\mathit{L}}^{\mathbf{+}}\left(y\right)\mathbf{=}\left(theleast-squaressolutionofL\left(\stackrel{\rightharpoonup }{x}\right)=\stackrel{\rightharpoonup }{y}\right)$.

1. Show that the transformation L⁺ is linear. Find the matrix A⁺ of L⁺ , in terms of the matrix A of L.
2. If L is invertible, what is the relationship between L⁺ and L^-1 ?
3. What is localid="1660646282368" ${\mathit{L}}^{\mathbf{+}}\mathbf{\left(}\mathbf{L}\left(\stackrel{\rightharpoonup }{x}\right)\mathbf{\right)}$, for $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$in localid="1662031720578">ℝn?
4. What is ${\mathit{L}}^{\mathbf{+}}\mathbf{\left(}\mathbf{L}\mathbf{\left(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{y}}\mathbf{\right)}\mathbf{\right)}$, for$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{y}}$in localid="1662031731206" ${\mathbf{ℝ}}^{\mathbf{m}}$?
5. Find L⁺ for the linear transformation $\mathit{L}\left(\stackrel{\rightharpoonup }{x}\right)\mathbf{=}\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$.

Short answer

1. It is proved that L⁺ is linear.
2. The relationship between L⁺ and L^-1 is L⁺ = L^-1.
3. The value of $L^{+}\left(L\left(\stackrel{\rightharpoonup }{x}\right)\right)=\stackrel{\rightharpoonup }{x}$.
4. The value of $L\left(L^{+}\left(\stackrel{\rightharpoonup }{y}\right)\right)=\stackrel{\rightharpoonup }{y}$.
5. The value of $L^{+}\left(\stackrel{\rightharpoonup }{y}\right)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\left(\stackrel{\rightharpoonup }{y}\right)$.

Step-by-step solution

Determine least Square solution

For a linear system $\mathit{A}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}\mathbf{=}\mathit{b}$, a vector${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}}^{\mathbf{+}}$in localid="1662034053845">ℝnis considered as least-squares solution of the given linear system if $||\stackrel{\rightharpoonup }{b}-A{\stackrel{\rightharpoonup }{x}}^{+}||\mathbf{\le }||\stackrel{\rightharpoonup }{b}-A\stackrel{\rightharpoonup }{x}||$for every$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$in localid="1662034074963">ℝm.

Prove that is linear

(a)

Given that $L\left(\stackrel{\rightharpoonup }{x}\right)=A\stackrel{\rightharpoonup }{x}$is a linear transformation from localid="1662034112751">ℝn→ℝm with . Let L⁺ is pseudo inverse of L such that $L^{+}=\left(theleast-aquraessolutionofL\left(\stackrel{\rightharpoonup }{x}\right)=\stackrel{\rightharpoonup }{y}\right)$.

Now, the least-squares solution of$L\left(\stackrel{\rightharpoonup }{x}\right)=\stackrel{\rightharpoonup }{y}$if$ker\left(L\right)=\left\{\stackrel{\rightharpoonup }{0}\right\}$is given by $\stackrel{\rightharpoonup }{x}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\rightharpoonup }{y}$. The least-squares solution of$L\left(\stackrel{\rightharpoonup }{x}\right)=\stackrel{\rightharpoonup }{y}$is given as $L^{+}\left(\stackrel{\rightharpoonup }{y}\right)$.

So, we have $L^{+}\left(\stackrel{\rightharpoonup }{y}\right)={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\rightharpoonup }{y}$.

Now, take $L^{+}\left(a\stackrel{\rightharpoonup }{y}+b\stackrel{\rightharpoonup }{z}\right)$as follows:

$\begin{array}{rcl}{L}^{+}\left(a\stackrel{\rightharpoonup }{y}+b\stackrel{\rightharpoonup }{z}\right)& =& \left(A^{T}A\right)A^T\left(a\stackrel{\rightharpoonup }{y}+b\stackrel{\rightharpoonup }{z}\right)\\ & =& a\left({\left(A^{T}A\right)}^{-1}{A}^T\right)\left(\stackrel{\rightharpoonup }{y}\right)+b\left({\left(A^{T}A\right)}^{-1}{A}^T\right)\left(\stackrel{\rightharpoonup }{z}\right)\\ & =& a{L}^{+}\left(\stackrel{\rightharpoonup }{y}\right)+b{L}^{+}\left(\stackrel{\rightharpoonup }{z}\right)\end{array}$

The above calculation shows that L⁺ is linear.

Now, we have $\begin{array}{rcl}{L}^{+}\left(\stackrel{\rightharpoonup }{y}\right)& =& {\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\rightharpoonup }{y}\\ & =& \left({\left(A^{T}A\right)}^{-1}{A}^T\right)\stackrel{\rightharpoonup }{y}\\ & =& A^{+}\stackrel{\rightharpoonup }{Y}\end{array}$. This can be manipulated as follows:

Thus, the value of A⁺ is ${\left(A^{T}A\right)}^{-1}{A}^T$.

Find relationship between L+ and L-1

(b)

It is given in the question that L is invertible because $ker\left(L\right)=\left\{\stackrel{\rightharpoonup }{0}\right\}$. Now, L⁺ is also invertible because L is invertible.

The above statement implies that if $L\left(\stackrel{\rightharpoonup }{x}\right)=\stackrel{\rightharpoonup }{y}$then $\stackrel{\rightharpoonup }{x}=L^{-1}\left(\stackrel{\rightharpoonup }{y}\right)$. From the calculation in part (a), we have $L^{+}\left(\stackrel{\rightharpoonup }{y}\right)=\stackrel{\rightharpoonup }{x}$.

Thus, the relationship between L⁺ and L^-1 is L⁺ = L^-1.

Find the value of L+(Lx⇀)

(c)

From part (a), we have $L^{+}\left(\stackrel{\rightharpoonup }{y}\right)={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\rightharpoonup }{y}$. So, the value of $L^{+}\left(L\left(\stackrel{\rightharpoonup }{x}\right)\right)$can be computed as follows:

$\begin{array}{rcl}{L}^{+}\left(L\left(\stackrel{\rightharpoonup }{x}\right)\right)& =& L^{+}\left(A\left(\stackrel{\rightharpoonup }{x}\right)\right)\\ & =& {\left(A^{T}A\right)}^{-1}{A}^T\left(A\left(\stackrel{\rightharpoonup }{x}\right)\right)\\ & =& {\left(A^{T}A\right)}^{-1}\left(A^{T}A\right)\left(\stackrel{\rightharpoonup }{x}\right)\\ & =& \stackrel{\rightharpoonup }{x}\end{array}$

Thus, the value of $\begin{array}{rcl}{L}^{+}\left(L\left(\stackrel{\rightharpoonup }{x}\right)\right)& =& \stackrel{\rightharpoonup }{x}\end{array}$.

Find the value of L(L+y⇀)

(d)

From part (a), we have $L^{+}\left(\stackrel{\rightharpoonup }{y}\right)={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\rightharpoonup }{y}$. The value of$L\left(L^{+}\left(\stackrel{\rightharpoonup }{y}\right)\right)$ can be obtained as follows:

$\begin{array}{rcl}L\left(L^{+}\left(y\right)\right)& =& L\left({\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\rightharpoonup }{y}\right)\\ & =& A\left({\left(A^{T}A\right)}^{-1}{A}^T\stackrel{\rightharpoonup }{y}\right)\\ & =& A\left(A^{-1}{A}^{T^{-1}}{A}^T\stackrel{\rightharpoonup }{y}\right)\\ & =& A{A}^{-1}{A}^T{A}^{T^{-1}}\stackrel{\rightharpoonup }{y}\\ & =& \stackrel{\rightharpoonup }{y}\end{array}$

Thus, the value of $\begin{array}{rcl}L\left(L^{+}\left(y\right)\right)& =& \stackrel{\rightharpoonup }{y}\end{array}$.

Find L+ for the given linear transformation

(e)

The given linear transformation is $L\left(\stackrel{\rightharpoonup }{x}\right)=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\stackrel{\rightharpoonup }{x}$. From part (a), we have$L^{+}\left(\stackrel{\rightharpoonup }{y}\right)=A^{+}\stackrel{\rightharpoonup }{y}$ where $A^{+}={\left(A^{T}A\right)}^{-1}{A}^T$.

Now, it is given $A=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$. Find the value of A⁺ using $A^{+}={\left(A^{T}A\right)}^{-1}{A}^T$. First find $A^{T}A$:

$\begin{array}{rcl}{A}^{T}A& =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\\ & =& \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

It is an identity matrix of 2x2 order. So, its inverse is also same which implies ${\left(A^{T}A\right)}^{-1}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.

Now, find the value of ${\left(A^{T}A\right)}^{-1}{A}^T$:

$\begin{array}{rcl}{\left(A^{T}A\right)}^{-1}{A}^T& =& \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\end{array}$

Thus, the required value is $L^{+}\left(\stackrel{\rightharpoonup }{y}\right)\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\end{array}\begin{array}{rcl}& & \left(\stackrel{\rightharpoonup }{y}\right)\end{array}$.